Design of Folded plate

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CE4210 1 Dr. Ataur Rahman, Prof. CE, KUET

Basic Concept of Folded Plate

Folded plates are sometimes called hipped plates. The principle was first used in Germany by Ehlers, in 1924, not for roofs but for large coal bunkers and he published a paper on the structural analysis in 1930. Then it was introduced to the United States of America

immediately after the Second World War where it rapidly became popular and accepted as a new form of construction because of its feasibility of erection, proven performance and its structural clarity on both analysis and design.

The distinguishing feature of the folded plate is the ease in forming plane surfaces. Therefore, they are more adaptable to smaller areas than curved surfaces which require multiple use of forms for maximum economy. A folded plate may be formed for about the same cost as a horizontal slab and has much less steel and concrete for the same spans. Folded plates are not adapted to as wide bay spacing as barrel vaults. For width of plate over, say, 12 feet, the thickness of the folded plate must be thicker than for a barrel vault. Some advantage may be gained by increasing the thickness of the slab just at the valleys so it will act as a hunched beam and as an I section plate girder.

Components of a Folded Plate roof

The principle components in a folded plate structure are illustrated in the sketch below. They consist of, 1) the inclined plates, 2) edge plates which must be used to stiffen the wide plates, 3) stiffeners to carry the loads to the supports and to hold the plates in line, and 4) columns to support the structure in the air. A strip across a folded plate is called a slab element because the plate is designed as a slab in that direction. The span of the structure is the greater distance between columns and the bay width is the distance between similar structural units.

The structure below is a two segment folded plate. If several units were placed side by side, the edge plates should be omitted except for the first and last plates. If the edge plate is not omitted on inside edges, the form should be called a two segment folded plate with a common edge plate.

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The folded plate structures may have a simple span, as shown below, or multiple spans of varying length, or the folded plate may cantilever from the supports without a stiffener at the end.

Fig. 1 Basic components of folded plate roof

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Fig. 2 Column supported folded plate roof structure

Fig. 3 Wall supported folded plate roof structure

Analysis Principles

Folded plate of the cross-section shown in figure below is very useful as it is simple to

construct, can span considerable distances and cover large or small areas. Its upstanding edge beams and internal valleys provide gutters for rainfall drainage and, if they are to be propped with columns, as in Fig. 2, the vertical edge beams are good for local structural, deflection and reinforcement reasons.

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The fundamental assumptions of folded plate

1. The material used (concrete) is homogeneous, monolithic and “elastic“.

2. Longitudinal vertices are continuous and held in such a way that there is no relative rotation or sliding along the common boundary between two adjoining plates.

3. Plates have very small resistance against twisting and torsion and therefore those stresses are usually considered small and thus ignored.

4. End diaphragms or bents, if any, are provided to transmit end reactions from the plates to the columns, however, they are generally considered unable to provide restraint against rotation at the end of the plates.

5. For longitudinal spans of 50 feet and over, bridging is recommended at midpoint. For spans from 70 to 100 feet, bridging should be provided at the third points of the span. Bridging should be of not less than 4 inches in thickness and have adequate openings at the bottom of the valley to facilitate drainage.

6. The principle of superposition holds true and valid; that means, the structure is susceptible to being analyzed separately for its redundant forces and different load conditions, then at the end the results are algebraically added to get to the final answer.

7. In all plates, the plane sections remain plane after deformation.

Loads on Folded Plate Roof 1) Self-weight

2) Superimposed dead load (normally 10-20 psf) 3) Live load ( 20 psf)

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4) Snow load (10-40 psf)

It is customary to consider all loads act at the ridge of the roof where two plates meet, thus it is called ridge load.

From the above figure,

_, ₁ cos ¹ _, ₁ cos and

sin sin

n n n n

n n n n

n n

P P

S  S 

 ^ 

   

If the plates were independent, they would develop different fiber stress, f_n and f_n+1 at the edge n. But the two plate being connected together, they cannot bend independently. Hence along the edge n these two stresses are same and shear stress tn develop along their common edge n.

So, it can be prove that,

Pn

n-2

n-1

Øn

n+1

n+2 n

Øn+1

γ

n

Sn,n-1 Sn,n+1

Mn Mn+1

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

1



1

1 1

4 _n _n _n

n n

T f f

A A_ ^

 

  

 

 

Now, consider the deflection of the plates,

Due to this deflection at the common edge n of these two plates, there will be rotation at the ridge. Where there is rotation there will be moment. So, a joint moment will be induce at this common edge n.

Thus the influence of joint moments can be accounted for by applying additional load at the ridge. This additional ridge load can be resolved into plate loads in the same manner P_n was resolved. So,

Pn

Δn

Δn-1

Δn+1

mn

Pn

mn-1

ΔPn

mn

mn+1

ln ln+1

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_, ₁ ^cos ¹ and _, ₁ ^cos

sin sin

n n n n

n n n n

n n

P P

S  S 

 ^ 

 

 

   

Where, ¹ ¹

1

n n n n

n

n n

m m m m

P l l

 



 

  

In the calculation of plate moment, plate deflection, edge shear and plate rotation, the total plate load caused by the ridge load Pn and the additional ridge load ΔPn have to be used. It is thus seen that the total load R_n on plate n is,

^Rⁿ 



^S^{n n}^, ¹^Sⁿ^1,ⁿ

 

 ^S^{n n}^, ¹ ^Sⁿ^1,ⁿ



Folded plate roof structures are analyzed by these three methods:

1) Whitney Method 2) Simpson Method 3) The Iteration method

Among these three, the Iteration method offers a simple mean of analyzing certain types of folded plate roof. Following example show the steps involved in Iteration method.

4.5”

15’ 15’ 7.5’

7.5’

3.5” 6’

0

1

2

3

60’

P-1 P-2

P-3

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Example and Sample Calculation:

1. Data Span = 60 ft

Self wt. of concrete = 150 pcf Live load = 15 psf of surface area

Thickness = 4.5 in for end plates and 3.5 in for inner plates.

From the plate geometry, it can be seen that the cross section of folded plate is repetitive after joint 3. So, the rest of the plates will have the same results of 1^st the three plates.

Plate

Plate width, h_n (ft)

Hz projection, l_n (ft)

Inclination to Hz Ø_n

Angle between plates, γ_n

A_n (ft²)

Z_n (ft³)

1 9.6 7.5 321.2 282.4 3.6 5.765

2 9.6 7.5 38.4 77.2 2.8 4.484

3 9.6 7.5 321.2 282.4 2.8 4.484

2. Load and fixed end moment (FEM)

Plate Load, lb/ft FEM (lb-ft/ft)

1 684 2565

2 564 352

3 564 352

For 4” cantilever plate,

FEM for plate-1: PL/2 = 684x7.5/2=2565 lb-ft/ft

For 3.5” plate, vertical load ^{3.5 150} ₁₅ ¹ ^58.75 ₇₅

12 cos cos(38.4)

w 

   

    

FEM for plate-2:

2 2

75 7.5

352 lb-ft/ft

12 12

wL 

  

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3. Moment Distribution

1 2 3

Distribution Factor 0 1 0.429 0.571 0.5 0.5 FEM -2566 +352 -352 +352 -352

+2213 -474 -632 -316

Final Moment= -2566 +2566 +280 -280 -668

Reaction for moment 380 380 126 126

Reaction for loads 684 282 282 282 282

Total Reaction = 1346 58 408 408

Loads (lb) 1346 58 817

The ridge loads obtained above and the resulting plate load are shown in the following figure.

Load on Plate = R/2sinθ, where θ is the inclination of the plate with the horizontal.

1346

58

817

1077

47

654 1124

700

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4. Bending Moment and Inplane flexural Stresses

For Plate-1,

Bending moment, M₁ = wL²/8 =484650 lb-ft

Bending stress, f=Mc/I= M/Z = 484650/5.765 = 84068 psf = 84068/144 = 584 psi 1

1 2

2 0

1

0

2

3

P-1

-584

+584 0

1

P-2

-783

+783 2

1

P-3

-488

+488 2

3

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5. Stress Distribution

0 1 2 3

Distribution Factor 7/16 9/16 0.5 0.5 0.5 Free Stress -584 +584 +783 -783 -488 +488

-43 +87 -112 +56

-59.5 +119 -119 +59.5 +13 -26

-1.8 +0.9

-33 -16.5 +8 -4

+2.2

-1.1

-8 +4

Final Stress= -613 +643 +643 -616 -616 +552

6. Plate Deflection Deflection at mid-span,

P-1

-613

+643 0

1

P-2

-616

+643 2

1

P-3

-616

+552 2

3

L w

h

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 

4

2

2 2

4 2

5 384

and 8

2 2

2 8 2 8

5 5

384 48

m

b t

m b t

wL EI

wL Mc

M f

I

Mc wL h wL h f f f

I I I

wL L

f f

EI Eh

 

   

 

 

 

      

 

   

      

   

For Plate 1, δ1 = -7064523/E ft For Plate 2, δ 2 = +7083066/E ft For Plate 3, δ 3 = -6574435/E ft 7. Calculate Joint Displacement

In Plate -2

1,2 1cos 2 1cot 2

7064523 7083066

1.025 0.225

8834826 =

w ec

E E

E

   

 

   

2,1 3cos 2 2cot 2

6574435 7083066

1.025 0.225

8332485

w ec

E E

E

   

 

   

 0

1 3

2

θ δ1

δ2

w1,2

θ

2θ

δ3

δ2

w2,1 2θ w2,3

δ3

2θ w3,2

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Transverse deflection in plate-2

 

2 1,2 2,1

8834826 8332485 502341

w w

E E

     

Transverse moment,

3

2 2

2 2 3

2

6 6 12 3.5 502341

67.6 ft-lb 9.6 12 12 12

EI E

h E

 

  

      

In plate-3

2,3 2cos 2 3cot 2

7083066 6574435

1.025 0.225

8739391 =

w ec

E E

E

   

 

   

3,2 3cot

6574435 8218043

1.25

w

E E

 



  

 

3 2,3 3,2

8739391 8218043 521348

w w

E E

     

Transverse moment,

3 3 3

2 2 3

3

6 6 12 3.5 521348

70.12 ft-lb 9.6 12 12 12

EI E

h E

 

  

      

8. Moment Distribution

1 2 3 Distribution Factor 0 1 3/7 4/7 0.5 0.5 FEM -67.6 -67.6

+67.6 +33.78 +44.5

-70.12 -70.12

+59.4 +29.7 Final Moment= 0 +10.75

10.75/7.5

-10.75 -40.43 51.2/7.5 Reaction due to moment 1.43 1.43 6.82 6.82

Total Reaction 1.43 8.26 6.82

Loads (lb) 1.43 8.26 6.82x2 13.64

(*These loads are parabolic load distribution)

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For Plate-1,

Bending moment, M₁ = wL²/π² =1.147x60²/3.14² =418.4 lb-ft

Bending stress, f=Mc/I= M/Z = 418.4/5.765 = 72.6 psf = 72.6/144 = 0.5 psi

9. Stress Distribution

0 1 2 3

Distribution Factor 7/16 9/16 0.5 0.5 0.5 Free Stress -584 -0.5 +0.5 -3.1 +3.1 -2.4 +2.4

+0.8 -1.6 +2.0 -1

+1.1 -2.2 +2.2 -1.1 +0.5

-0.25

-0.6 +0.3 -0.15 +0.075

+0.15 -0.075

Final Stress= +0.017 -0.54 -0.54 -0.01 -0.01 +1.26 1.43

8.26

13.64

1.147

6.61

10.92 5.461

4.314

P-1

-0.5

+0.5 0

1

P-2

+3.1

-3.1 2

1

P-3

-2.4

+2.4 2

3

P-1

+0.017

-0.54 0

1

P-2

-0.01

-0.54 2

1

P-3

-0.01

+1.26 2

3

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10.Final Stresses

It is observed that these stresses are small enough compared with those obtained from the moment due to self-wt and live load. The final stresses are obtained by adding these two stresses i.e stress due to load and stress due to deflection.

+

=

11.Reinforcement detailing Material Properties:

Concrete: 30 MPa (4350 psi)

Allowable stress in concrete, _f_c __0.25_f_c^' __{0.25 4350}_ _₁₀₈₈_psi P-1

-613

+643 0

1

P-2

-616

+643 2

1

P-3

-616

+552 2

3

P-1

+0.017

-0.54 0

1

P-2

-0.01

-0.54 2

1

P-3

-0.01

+1.26 2

3

P-1

-613

+643 0

1

P-2

-616

+643 2

1

P-3

-616

+553 2

3

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Steel: 412 MPa (60 ksi)

Allowable stress in steel, _f_s __0.4_f_y __{0.4 60}_ _₂₄_ksi

From the above stress diagrams, it is evident that no steel is needed at the compression side of the plate.

Steel for tension zone:

For plate-1

Longitudinal tension force= ¹ ¹ ₆₄₃ ^{9.6 12 1.5} _4.5 _82.2

2 ^t 2 2 2

T  f    d t      kips

Required steel, ^82.2 _{3.42 in /}²

s 24

s

A T ft

 f  

Provide 4 bars of 25mm Ø.

Bar Size (mm) Area (sq.in) bar no. area (in2)

8 0.077 #2 0.05 10 0.12 #3 0.11

12 0.175 #4 0.20 16 0.31 #5 0.31 20 0.48 #6 0.44

22 0.58 #7 0.60

25 0.785 #8 0.79 28.7 1.00 #9 1.00 32.3 1.27 #10 1.27

Plate-2

Longitudinal tension force= ¹ ¹ 643 ^{9.6 12 1.5} 3.5 64

2 ^t 2 2 2

T f d t   kips

       

Required steel, ⁶⁴ 2.66 in /²

s 24

s

A T ft

 f  

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Provide 2 bars of 25mm Ø and 2 bars of 22 mm Ø

For plate-3

Longitudinal tension force= ¹ ¹ 553 ^{9.6 12 1.5} 3.5 55

2 ^t 2 2 2

T  f    d t      kips

Required steel, ⁵⁵ 2.3 in /²

s 24

s

A T ft

 f  

Provide 4 bars of 22mm Ø.

4- 25mm Ø

2- 25mm Ø 2-22 mm Ø

4- 22mm Ø 12 mm Ø binder @ 16” c/c